Qn: Sulfur dioxide causes a colour change in acidified potassium dichromate. Given that the potassium dichromate and sulfur dioxide are present in the reacting ratio of 1:3, suggest a likely sulfur−containing product for the reaction.
This is considered quite a common redox question.
Firstly, what do you need to know? And how to apply what you know to solve this question?
Basic knowledge of common oxidizing and reducing agents. Look up your lecture notes, there should be a table of common O.A. and R.A. together with their respective products. This will allow you to identify which is the oxidizing/reducing agent.
You will also need to calculating the reacting mole ratio between the O.A. and R.A. However, in this particular question, the reacting ratio is given.
From the Cr half-equation (you will need to know how to balance half-equations in case they are not found in data booklet), you can determine the number of electrons gained for Cr since you know the product formed. (Note that you will not be able to do the same for sulfur dioxide as you do not know the product form)
Electron transfer is the main concept in Redox.
All electrons lost are equal to the electrons gained.
Once we know 6 electrons are gained by the two Chromium atoms, we can deduce that the three sulfur atoms lost 6 electrons.
Therefore, each sulfur atom would have lost 2 electrons and gained +2 oxidation state.
Since sulfur in sulfur dioxide has an oxidation state of +4, the sulfur in the new sulfur-containing product will have an oxidation state of +6, i.e. sulfur trioxide or sulfate ion.